Binomial expansion for rational n and series modelling – Year 13 A level maths worksheet answers to download as PDF

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1. Answer to Question 1

   Method: use \((1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots\), replacing \(x\) by the inside term when needed, and remember that the series is only valid when the inside term has modulus less than 1.

   1. Here \(n\) is a negative integer, so the series never stops. \(n=-2\): coefficients \(-2\), \(\dfrac{(-2)(-3)}{2!}=3\), \(\dfrac{(-2)(-3)(-4)}{3!}=-4\). So \((1+x)^{-2}=1-2x+3x^2-4x^3+\cdots\), valid for \(|x|<1\). The coefficients alternate in sign because \(n\) is negative.
   2. Writing \(1-x\) as \(1+u\) avoids sign errors. Let \(u=-x\) and use \((1+u)^{-3}=1-3u+6u^2-10u^3+\cdots\) (coefficients \(-3\), \(\dfrac{(-3)(-4)}2=6\), \(\dfrac{(-3)(-4)(-5)}6=-10\)). Then \((1-x)^{-3}=1+3x+6x^2+10x^3+\cdots\), valid for \(|x|<1\). Every coefficient is now positive, since \(u=-x\) changes each odd power.
   3. This is a geometric series with ratio \(-2x\). Let \(u=2x\): \((1+u)^{-1}=1-u+u^2-u^3+\cdots\). So \((1+2x)^{-1}=1-2x+4x^2-8x^3+\cdots\). It is valid when \(|2x|<1\), that is \(|x|\lt \tfrac12\). The range of validity is narrower than in parts a) and b).

2. Answer to Question 2

   Method: with a fractional power the expansion is infinite. Work out each coefficient carefully as a product of fractions, then substitute the inside term and simplify.

   1. Take \(n=\tfrac13\). Coefficients: \(\dfrac13\), \(\dfrac{\frac13\cdot(-\frac23)}{2}=-\dfrac19\), \(\dfrac{\frac13(-\frac23)(-\frac53)}{6}=\dfrac5{81}\). So \((1+x)^{1/3}=1+\dfrac x3-\dfrac{x^2}9+\dfrac{5x^3}{81}-\cdots\), valid for \(|x|<1\). So the first four terms are \(1\), \(\tfrac x3\), \(-\tfrac{x^2}9\) and \(\tfrac{5x^3}{81}\).
   2. Take \(n=-\tfrac12\) and replace \(u\) by \(-x\). \((1+u)^{-1/2}=1-\dfrac u2+\dfrac{3u^2}8-\dfrac{5u^3}{16}+\cdots\) with \(u=-x\): \((1-x)^{-1/2}=1+\dfrac x2+\dfrac{3x^2}8+\dfrac{5x^3}{16}+\cdots\), valid for \(|x|<1\). All the terms are positive because the minus signs cancel in pairs.
   3. Write \(\sqrt{1+4x}=(1+4x)^{1/2}\). \((1+u)^{1/2}=1+\dfrac u2-\dfrac{u^2}8+\dfrac{u^3}{16}-\cdots\) with \(u=4x\): \(1+2x-2x^2+4x^3-\cdots\). Valid when \(|4x|<1\), so \(|x|\lt \tfrac14\). Notice that each power of \(4x\) must be calculated in full: \((4x)^2=16x^2\) and \((4x)^3=64x^3\).

3. Answer to Question 3

   Method: the binomial series needs the form \((1+\text{something})^n\). So first take out the constant term as a factor, raised to the power \(n\), then expand and multiply back.

   1. Take out \(4\) from the bracket, which gives the factor \(4^{-1/2}=\tfrac12\). \((4+x)^{-1/2}=4^{-1/2}\left(1+\dfrac x4\right)^{-1/2}=\dfrac12\left(1-\dfrac x8+\dfrac38\cdot\dfrac{x^2}{16}-\cdots\right)=\dfrac12-\dfrac x{16}+\dfrac{3x^2}{256}-\cdots\). Valid for \(\left|\dfrac x4\right|<1\), so \(|x|<4\). So the expansion is \(\tfrac12-\tfrac{x}{16}+\tfrac{3x^2}{256}\), valid for \(-4\lt x<4\).
   2. Take out \(2\), which gives the factor \(2^{-2}=\tfrac14\). \((2+3x)^{-2}=\dfrac14\left(1+\dfrac{3x}2\right)^{-2}=\dfrac14\left(1-3x+\dfrac{27x^2}4-\cdots\right)=\dfrac14-\dfrac{3x}4+\dfrac{27x^2}{16}-\cdots\) using \((1+u)^{-2}=1-2u+3u^2\) with \(u=\dfrac{3x}2\). Valid for \(|x|\lt \tfrac23\). The validity comes from \(\left|\tfrac{3x}{2}\right|<1\).
   3. Write the fraction as \((3-2x)^{-1}\) and take out \(3\). \((3-2x)^{-1}=\dfrac13\left(1-\dfrac{2x}3\right)^{-1}=\dfrac13\left(1+\dfrac{2x}3+\dfrac{4x^2}9+\cdots\right)=\dfrac13+\dfrac{2x}9+\dfrac{4x^2}{27}+\cdots\). Valid for \(|x|\lt \tfrac32\). The validity comes from \(\left|\tfrac{2x}{3}\right|<1\), which gives \(|x|\lt \tfrac32\).

4. Answer to Question 4

   Method: you do not need the whole expansion. Find only the terms that can produce \(x^3\), and when a bracket multiplies a series, collect every product whose powers add up to 3.

   1. The \(x^3\) term is \(\dfrac{n(n-1)(n-2)}{3!}(-3x)^3\) with \(n=-\tfrac13\). Coefficient \(=\dfrac{(-\frac13)(-\frac43)(-\frac73)}{3!}(-3)^3=\dfrac{-28/27}{6}\times(-27)=\dfrac{28}6=\dfrac{14}3\). The coefficient of \(x^3\) is \(\tfrac{14}{3}\).
   2. Expand \((1+x)^{-1}\) first, then multiply by each term of \(2-x\). \((1+x)^{-1}=1-x+x^2-x^3+\cdots\). Multiplying by \((2-x)\), the \(x^3\) terms are \(2(-x^3)+(-x)(x^2)=-2x^3-x^3=-3x^3\). Coefficient \(=-3\). The two contributions come from \(2\times(-x^3)\) and \((-x)\times x^2\).
   3. Write the expression as \((1+x)(1-x)^{-1/2}\). \((1-x)^{-1/2}=1+\dfrac x2+\dfrac{3x^2}8+\dfrac{5x^3}{16}+\cdots\). Multiplying by \((1+x)\): \(x^3\) terms are \(\dfrac5{16}+\dfrac38=\dfrac{11}{16}\). The \(x^3\) terms come from \(1\times\tfrac{5}{16}x^3\) and \(x\times\tfrac38x^2\), so the coefficient is \(\tfrac{11}{16}\).

5. Answer to Question 5

   Method: use \((1+x)^{1/2}\approx1+\dfrac x2-\dfrac{x^2}8\). Choose \(x\) so that \(1+x\) is the number under the square root, and check that \(|x|<1\) so the series converges.

   1. Here \(1.04=1+0.04\). \(x=0.04\): \(1+0.02-\dfrac{0.0016}8=1+0.02-0.0002=1.0198\). Calculator: 1.019804. The error is about \(4\times10^{-6}\). The estimate is very accurate because \(x\) is small.
   2. Here \(0.98=1-0.02\), so \(x\) is negative. \(x=-0.02\): \(1-0.01-\dfrac{0.0004}8=1-0.01-0.00005=0.98995\). Calculator: 0.989949. The error is about \(5\times10^{-7}\). The estimate is even better because \(|x|\) is smaller than in part a).
   3. We cannot use \(x=49\) because the series would diverge, so we factor out the perfect square \(49\). \(\sqrt{50}=7\sqrt{1+\tfrac1{49}}\approx7\left(1+\dfrac1{98}-\dfrac1{8\times2401}\right)=7\times1.0101523=7.071066\). Calculator: 7.071068, so the estimate is correct to 5 decimal places. This shows why factoring out the nearest perfect square is such a powerful trick.

6. Answer to Question 6

   Method: the expansion of \((1+2x)^n\) is only valid when the inside term \(2x\) has modulus less than 1. Always check this before substituting a value.

   1. The condition for validity is that the term added to 1 has modulus less than 1. \(|2x|<1\) so \(|x|\lt \tfrac12\). So the expansion is valid for \(-\tfrac12\lt x\lt \tfrac12\).
   2. Check the condition with the chosen value. \(x=1\) is outside the range \(|x|\lt \tfrac12\), so the series does not converge to \(\dfrac1{\sqrt3}\) (the terms \(1-x+\tfrac32x^2-\cdots\) get larger, not smaller). So substituting \(x=1\) gives a meaningless answer.
   3. Find the coefficients with \(n=-\tfrac12\) and inside term \(2x\). \((1+2x)^{-1/2}=1-x+\dfrac{3}{2}x^2-\cdots\) (using \(-\tfrac12\), then \(\dfrac{(-\frac12)(-\frac32)}2(2x)^2=\tfrac32x^2\)). With \(x=0.01\): \(1-0.01+0.00015=0.99015\). Calculator: 0.990148, so 0.99015 to 5 d.p. The value \(x=0.01\) is inside the range of validity, so the estimate can be trusted.

7. Answer to Question 7

   1. Replace \(L\) by \(1.03L_0\) in the formula and separate the square root. \(T=2\pi\sqrt{\dfrac{1.03L_0}g}=\sqrt{1.03}\times2\pi\sqrt{\dfrac{L_0}g}=T_0(1.03)^{1/2}\). This proves the result.
   2. Use \((1+x)^{1/2}\approx1+\tfrac x2-\tfrac{x^2}8\) with \(x=0.03\). \(x=0.03\): \(1+\dfrac{0.03}2-\dfrac{0.0009}8=1+0.015-0.0001125=1.0148875\). So \((1.03)^{1/2}\approx\) 1.01489 (calculator 1.014889). The three-term estimate agrees with the calculator to 5 decimal places.
   3. A longer period means each tick takes longer, so the clock runs slow. Each swing now takes \(2\times1.0148875=2.029775\) s. In one real day of 86 400 s the clock shows \(86400\div1.0148875=85\,134\) s. It loses \(86400-85134\approx\) 1267 s (about 21 minutes). So the clock loses about 1267 seconds per day, which is roughly 21 minutes.

8. Answer to Question 8

   1. Write \(\gamma=(1-u)^{-1/2}\) with \(u=\tfrac{v^2}{c^2}\) and use \(n=-\tfrac12\). Let \(u=\dfrac{v^2}{c^2}\). \((1-u)^{-1/2}=1+\dfrac u2+\dfrac{3u^2}8+\cdots\), so \(\gamma\approx1+\dfrac{v^2}{2c^2}+\dfrac{3v^4}{8c^4}\). This expansion is valid because \(v\lt c\), so \(0\le u<1\).
   2. Substitute the expansion of \(\gamma\) into \(E=m_0c^2(\gamma-1)\). \(E=m_0c^2(\gamma-1)\approx m_0c^2\left(\dfrac{v^2}{2c^2}+\cdots\right)=\tfrac12m_0v^2+\dfrac{3m_0v^4}{8c^2}\). For small \(v\) the second term is tiny, so \(E\approx\tfrac12m_0v^2\), the classical kinetic energy. This shows that Newton's formula is the first term of the relativistic formula.
   3. With \(v=0.1c\), \(u=0.1^2=0.01\). \(u=0.01\): \(\gamma\approx1+0.005+\dfrac{3(0.0001)}8=1+0.005+0.0000375=\) 1.0050375 (exact 1.0050378). The estimate differs from the exact value only in the seventh decimal place.

9. Answer to Question 9

   1. Write the new volume as 1.05 times the old one and cancel \(\tfrac43\pi\). \(\dfrac43\pi r_1^3=1.05\times\dfrac43\pi r_0^3\) gives \(r_1=r_0(1.05)^{1/3}\). Taking the cube root of both sides proves the result.
   2. Use \(n=\tfrac13\) and \(x=0.05\). \((1+x)^{1/3}=1+\dfrac x3-\dfrac{x^2}9\); with \(x=0.05\): \(1+0.016667-0.000278=1.016389\), so 1.0164 to 4 d.p. So the radius increases by about 1.64%.
   3. Surface area depends on \(r^2\), so it is multiplied by the square of the radius factor. Area is multiplied by \((1.05)^{2/3}=(1.016389)^2=1.03305\). The surface area increases by about 3.3%. So the surface area rises by about 3.3%, less than the 5% rise in volume.

10. Answer to Question 10

    1. Replace \(d\) by \(1.05d\) in the formula. \(I=\dfrac k{(1.05d)^2}=\dfrac k{d^2}(1.05)^{-2}\). So the intensity is multiplied by \((1.05)^{-2}\).
    2. Use \(n=-2\) and \(x=0.05\). \((1+x)^{-2}=1-2x+3x^2-4x^3\): \(1-0.1+0.0075-0.0005=0.907\). Calculator: 0.907029, so 0.9070 to 4 d.p. Four terms already give 4 decimal places of accuracy.
    3. Subtract the multiplier from 1: \(1-0.907=0.093\). The intensity becomes 0.907 of its original value, a fall of about 9.3%. So moving the lamp only 5% further away reduces the light by almost twice that percentage.

11. Answer to Question 11

    1. Let \(x=\tfrac hR\) and use \((1+x)^{-2}\approx1-2x\). \(x=\dfrac{10}{6371}=0.0015696\). \(g\approx9.81(1-2x)=9.81\times0.996861=\) 9.779 m s\(^{-2}\). So at cruising height \(g\) is about 9.779 m s\(^{-2}\).
    2. The relative change is \(-2x\), so convert \(2x\) into a percentage. \(2x=0.31392\%\), so the decrease is about 0.31%. So gravity is only about 0.31% weaker at 10 km.
    3. State the condition for the series and compare it with the given height. The expansion needs \(\left|\dfrac hR\right|<1\), that is \(h<6371\) km. Since 10 km is far smaller, the series converges very quickly. So the two-term approximation is both valid and accurate.

12. Answer to Question 12

    1. Use \((1+x)^{-2}\approx1-2x+3x^2\) with \(x=0.03\), then multiply by 5000. \(x=0.03\): \((1.03)^{-2}\approx1-0.06+3(0.0009)=0.9427\). Present value \(\approx5000\times0.9427=\) £4713.50. So the estimated present value is £4713.50.
    2. Compute \((1.03)^2=1.0609\) and divide. \(5000\div1.0609=\) £4712.98. This is the exact present value.
    3. Subtract the exact value from the estimate. \(4713.50-4712.98=\) £0.52. So the binomial estimate is only 52p too high, an error of about 0.01%.

13. Answer to Question 13

    1. Let the speed be multiplied by \(a\) and compare the new power with 1.08 times the old power. \(k(av)^3=1.08kv^3\) gives \(a^3=1.08\), so \(a=(1.08)^{1/3}\). So the speed must rise by a factor of \((1.08)^{1/3}\).
    2. Use \((1+x)^{1/3}\approx1+\tfrac x3-\tfrac{x^2}9\) with \(x=0.08\). \(x=0.08\): \(1+\dfrac{0.08}3-\dfrac{0.0064}9=1+0.026667-0.000711=1.025956\). So 1.0260 to 4 d.p. (calculator 1.025986). The three-term estimate is about 0.003% below the true value.
    3. Convert the factor into a percentage increase. The speed rises by about 2.6%. So an 8% rise in power needs only about a 2.6% rise in wind speed.

14. Answer to Question 14

    1. Factor out the perfect square 49 inside the root. \(7^2=49\), so \(\sqrt{50}=\sqrt{49\times\tfrac{50}{49}}=7\sqrt{1+\tfrac1{49}}\). This puts the root in the form \(7(1+x)^{1/2}\) with \(x=\tfrac1{49}\).
    2. Use the first two terms \(1+\tfrac x2\) and multiply by 7. \(7\left(1+\dfrac1{98}\right)=7+\dfrac1{14}=7.0714\) m. So the two-term estimate of the diagonal is about 7.0714 m.
    3. Add the third term \(-\tfrac{x^2}8\) and compare both estimates with the exact value. Three terms: \(7\left(1+\dfrac1{98}-\dfrac1{19208}\right)=7.0710641\). Exact \(\sqrt{50}=7.0710678\). Two-term error: \(\dfrac{7.0714286-7.0710678}{7.0710678}=0.0051\%\). So the two-term estimate is already within about 0.005% of the true length.

15. Answer to Question 15

    1. Compare the general coefficients \(na\) and \(\tfrac{n(n-1)}{2}a^2\) with 6 and 24. Coefficient of \(x\): \(na=6\). Coefficient of \(x^2\): \(\dfrac{n(n-1)}2a^2=24\), so \(n(n-1)a^2=48\). Substituting \(na=6\): \(6(n-1)a=48\), so \((n-1)a=8\). These are the two required equations.
    2. Subtract the two linear equations to eliminate \(n\). Subtracting: \(na-(n-1)a=a=6-8=-2\). Then \(n=\dfrac6a=-3\). So \(a=-2\), \(n=-3\). Check: \(na=(-3)(-2)=6\) and \((n-1)a=(-4)(-2)=8\).
    3. Use \(\tfrac{n(n-1)(n-2)}{3!}a^3\). \(\dfrac{(-3)(-4)(-5)}{6}(-2)^3=(-10)(-8)=\) 80. So the coefficient of \(x^3\) is 80.
    4. The inside term is \(-2x\). \(|-2x|<1\), so \(|x|\lt \tfrac12\). So the expansion is valid for \(-\tfrac12\lt x\lt \tfrac12\).
    5. Evaluate the exact value and the four-term series. LHS \(=(1-0.02)^{-3}=1.06248\ldots\) and RHS \(=1+0.06+0.0024+0.00008=1.06248\). They agree to 5 decimal places. The agreement confirms the values of \(a\) and \(n\).

16. Answer to Question 16

    1. Multiply both sides by the denominator and substitute values of \(x\) that remove one constant each time. \(4-5x=A(2-x)+B(1+x)\). Put \(x=-1\): \(9=3A\) so \(A=3\). Put \(x=2\): \(-6=3B\) so \(B=-2\). So \(f(x)=\dfrac3{1+x}-\dfrac2{2-x}\).
    2. Expand each fraction separately with \(n=-1\), taking out the factor 2 in the second one. \(\dfrac3{1+x}=3(1-x+x^2-x^3+\cdots)\). \(\dfrac{-2}{2-x}=-\left(1-\dfrac x2\right)^{-1}=-\left(1+\dfrac x2+\dfrac{x^2}4+\dfrac{x^3}8+\cdots\right)\). Adding: constant \(3-1=2\); \(x\): \(-3-\dfrac12=-\dfrac72\); \(x^2\): \(3-\dfrac14=\dfrac{11}4\); \(x^3\): \(-3-\dfrac18=-\dfrac{25}8\). Collecting like terms gives the required series.
    3. Each fraction has its own range of validity; the whole series needs both. The first part needs \(|x|<1\) and the second needs \(\left|\dfrac x2\right|<1\), i.e. \(|x|<2\). Both hold when \(|x|\lt \) 1. So the expansion is valid for \(-1\lt x<1\).
    4. Substitute \(x=0.1\) into the fraction and into four terms of the series. Exact: \(f(0.1)=\dfrac{3.5}{1.1\times1.9}=1.674641\). Series: \(2-0.35+0.0275-0.003125=1.674375\). Error: \(\dfrac{0.000266}{1.674641}\approx0.016\%\). So the percentage error is only about 0.016%.

17. Answer to Question 17

    1. \(A_3=1+0.25-\dfrac{0.25}8+\dfrac{0.125}{16}=1+0.25-0.03125+0.0078125=1.2266\). The exact value is \(\sqrt{1.5}=1.2247\).
    2. For \(x=0.1\) the values 1.0500, 1.0488, 1.0488 move quickly to the exact value 1.0488. For \(x=0.5\) the values 1.2500, 1.2188, 1.2266 swing either side of 1.2247 and close in more slowly, so more terms are needed for the same accuracy.
    3. For \(x=2\) the values are 2.0000, 1.5000, 2.0000 while the exact value is \(\sqrt3=1.7321\). The sequence does not settle. The expansion of \((1+x)^{1/2}\) is only valid for \(|x|<1\) and \(x=2\) is outside this range.
    4. \(A_2=1+0.05-0.00125=1.04875\). Exact \(=1.048809\). Error \(=\dfrac{1.048809-1.04875}{1.048809}\approx0.0056\%\).

18. Answer to Question 18

    1. \(\left(1-\dfrac{v^2}{c^2}\right)^{-1/2}\approx1+\dfrac{v^2}{2c^2}+\dfrac{3v^4}{8c^4}\), so \(E=m_0c^2(\gamma-1)\approx\tfrac12m_0v^2+\dfrac{3m_0v^4}{8c^2}\).
    2. \(v=0.2c\): \(\gamma=(1-0.04)^{-1/2}=1.0206207\), so \(E=0.0206207\,m_0c^2\). Classical value \(=\tfrac12(0.04)=0.02\,m_0c^2\). Difference \(=0.0006207\), which is \(\dfrac{0.0006207}{0.0206207}=\) 3.0% of \(E\).
    3. Run C is least accurate. \(v=0.5c\): \(\gamma=(0.75)^{-1/2}=1.154701\), \(E=0.154701\,m_0c^2\) while the classical value is \(0.125\,m_0c^2\), so the classical formula is \(19\%\) too low. The expansion needs \(\dfrac{v^2}{c^2}\) small, and \(0.25\) is not small.

19. Answer to Question 19

    1. Expand \((1-2x)^{-1/2}\) with inside term \(-2x\), then multiply by \((1+x)\) and collect terms. \((1-2x)^{-1/2}=1+\dfrac12(2x)+\dfrac{3}{8}(2x)^2+\dfrac5{16}(2x)^3+\cdots=1+x+\dfrac32x^2+\dfrac52x^3+\cdots\). Multiplying by \((1+x)\): \(1+(1+1)x+\left(\dfrac32+1\right)x^2+\left(\dfrac52+\dfrac32\right)x^3=1+2x+\dfrac52x^2+4x^3+\cdots\). So \(g(x)=1+2x+\tfrac52x^2+4x^3+\cdots\).
    2. The validity only depends on the bracket with a non-integer power. \(|2x|<1\), so \(|x|\lt \tfrac12\). So the expansion is valid for \(-\tfrac12\lt x\lt \tfrac12\).
    3. Substitute \(x=0.01\) into the four terms. \(1+0.02+0.00025+0.000004=1.020254\). Calculator: 1.020254. So the estimate is 1.020254, correct to 6 decimal places.
    4. Replace \(g(x)\) by its quadratic approximation and solve with the formula. \(1+2x+\tfrac52x^2=1.05\) gives \(5x^2+4x-0.1=0\), so \(x=\dfrac{-4+\sqrt{16+2}}{10}=\dfrac{-4+4.2426}{10}=\) 0.0243 (taking the small positive root). So the small root is approximately \(x=0.0243\).

20. Answer to Question 20

    1. \((1+u)^{-1/2}=1-\dfrac u2+\dfrac38u^2-\dfrac5{16}u^3+\dfrac{35}{128}u^4-\cdots\). With \(u=-4x\): \(1+2x+\dfrac38(16x^2)+\dfrac5{16}(64x^3)+\dfrac{35}{128}(256x^4)=1+2x+6x^2+20x^3+70x^4+\cdots\).
    2. \(\dbinom21=2\), \(\dbinom42=6\), \(\dbinom63=20\), \(\dbinom84=70\). All match.
    3. \(x=0.01\): \(1+0.02+0.0006+0.00002=1.02062\) (the next term is \(0.0000007\)). Calculator: \(0.96^{-1/2}=1.020621\), so 1.02062 to 5 d.p.
    4. \(x=\tfrac14\) gives \(|4x|=1\), which is not less than 1, so the series (with coefficients \(\binom{2n}n\to\infty\)) diverges; the function itself is undefined at \(x=\tfrac14\).